arguments in complex numbers

Question

I need to show that if $\arg\left(\frac{z_1+z_2}{z_1-z_2}\right)=\frac{\pi}{2}$ then $|z_1|=|z_2|$. How should I work this out? I know that $\arg\left(\frac{z_1+z_2}{z_1-z_2}\right) = \arg(z_1+z_2)-\arg(z_1-z_2)$ and since the argument is $\frac{\pi}{2}$ than the complex number lies on the positive imaginary axis. Is this correct?

Answer 1

Yes, this is correct.

Let $z_1=u+iv$, $z_2=r+is$ and equal the real part of the quotient to $0$: you'll be done in no time.

Answer 2

$T(z) = (1+z)/(1-z)$ is a Möbius transformation with $T(-1) = 0$, $T(1) = \infty$ and $T(i) = i$ and therefore mapping the unit circle onto the (extended) imaginary axis.

So $ \arg T(z) = \pi/2 $ implies $ |z| = 1 $. Now set $ z = z_2/z_1$.

(A possible geometric argument: $ \arg(1+z) - \arg(1-z) = \pi/2$ implies that the triangle with the points $(1, -1, z)$ has a right angle at $z$. Using the Converse of Thales Theorem if follows that $z$ lies on the circle whose diameter is the hypothenuse $[-1, 1]$.)